Problem: There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

<image>

You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MAX = 3 * 100 * 1000;
const double INF = 1e50;
int n;
double sy, sl, sr;
struct {
  double l, r;
} segments[MAX];
double prefix_length[MAX];
double x_intercept(double x1, double y1, double x2, double y2) {
  assert(y1 < 0 && 0 < y2);
  return x1 + ((x2 - x1) / (y2 - y1)) * (-1 * y1);
}
int find_segment(double v) {
  if (v > segments[n - 1].r) return n;
  int l = 0, r = n - 1;
  while (l < r) {
    const int m = (l + r) / 2;
    if (segments[m].r < v)
      l = m + 1;
    else
      r = m;
  }
  return l;
}
int main(void) {
  scanf("%lf %lf %lf %d", &sy, &sl, &sr, &n);
  for (int i = 0; i < n; i++) {
    scanf("%lf %lf", &segments[i].l, &segments[i].r);
    prefix_length[i] = segments[i].r - segments[i].l;
    if (i > 0) prefix_length[i] += prefix_length[i - 1];
  }
  int q;
  for (scanf("%d", &q); q--;) {
    double x, y;
    scanf("%lf %lf", &x, &y);
    const double multiplier = (y - sy) / y;
    double answer = 0;
    const double intial_xi = x_intercept(sl, sy, x, y);
    const double final_xi = x_intercept(sr, sy, x, y);
    int first = find_segment(intial_xi);
    int last = find_segment(final_xi);
    if (first != n) {
      if (segments[first].l < intial_xi) {
        answer += min(final_xi, segments[first].r) - intial_xi;
        first++;
      }
      if (last == n) {
        last--;
      } else if (final_xi < segments[last].r && first <= last) {
        answer += max(0.0, final_xi - segments[last].l);
        last--;
      }
      if (first <= last) {
        assert(intial_xi <= segments[first].l);
        assert(final_xi >= segments[last].r);
        answer +=
            prefix_length[last] - ((first == 0) ? 0 : prefix_length[first - 1]);
      }
    }
    printf("%lf\n", multiplier * answer);
  }
  return 0;
}